The steady-state response of a network to the excitation V cos(ωt + ϕ) may be found in three steps. The first two steps are as follows:

1. Determining the response of the network to the excitation e^jωt

2. Multiplying the above response by V̅ = Ve^jϕ 

The third step is

Concept:

• The same form of the sinusoidal signal is produced at the output.
• The only change will be in magnitude and phase.

The following block diagram shows the output

h(t) is the impulse response of the LTI system.

f r(t) = A sin (ωot), and the open loop transfer function be G(s) = G(jωo) = |G(jωo)| ∠G(jωo)

The output signal c(t) can be expressed as:

c(t) = A |G(jωo)|⋅ sin (ωot + ∠G(jωo))

Analysis:

Input = V cos(ωt + ϕ) 

Step 1:

e^jω = [cos (ωt) + sin (ωt)]

Step 2:

Multiplying the above input with Ve^jϕ , we will get

Ve^jϕ e^jω = V[cos ϕ + j sin ϕ] [cos (ωt) + sin (ωt)]

= V[ cos ϕ cos ωt + cos ϕ sin ωt + j sin ϕ cos ωt + j sin ϕ sin ωt] ---(1)

Step 3:

The real part of the equation (1), will give the response, since output is cosine signal.